How do you find all the asymptotes for function #f(x)= (x+3)/(x^2-9)#?

1 Answer
Jul 19, 2015

Answer:

#f(x) = (x+3)/(x^2-9) = (x+3)/((x-3)(x+3)) = 1/(x-3)#

with exclusion #x != -3#.

Hence the asymptotes are #y = 0# and #x = 3#.

Explanation:

Use the difference of squares identity: #a^2-b^2 = (a-b)(a+b)#

with #a=x# and #b=3# to find:

#(x^2-9) = (x^2-3^2) = (x-3)(x+3)#

So:

#f(x) = (x+3)/(x^2-9) = (x+3)/((x-3)(x+3)) = 1/(x-3)#

with exclusion #x != -3#.

As #x->+-oo# we find #f(x) -> 0#, so #y=0# is a horizontal asymptote.

There is a vertical asymptote at #x=3# since the denominator becomes zero and the numerator is non-zero.